Solved on Jan 18, 2024

Find the zeros of the quadratic relation $y=-3(x-5)(x+7)$ and determine if the parabola opens upward or downward.

STEP 1

Assumptions
1. The quadratic relation is given by $y=-3(x-5)(x+7)$ .
2. Zeros of the relation are the values of $x$ for which $y=0$ .
3. The direction in which the parabola opens is determined by the sign of the coefficient of the $x^2$ term.

STEP 2

To find the zeros of the relation, we set $y$ to zero and solve for $x$ .
$0 = -3(x-5)(x+7)$

STEP 3

Since the product of the two factors is zero, we can set each factor equal to zero and solve for $x$ .
$(x-5) = 0 \quad \text{or} \quad (x+7) = 0$

STEP 4

Solve the first equation for $x$ .
$x - 5 = 0 \Rightarrow x = 5$

STEP 5

Solve the second equation for $x$ .
$x + 7 = 0 \Rightarrow x = -7$

STEP 6

The zeros of the relation are $x = 5$ and $x = -7$ .

STEP 7

To determine the direction in which the parabola opens, we look at the coefficient of the $x^2$ term in the quadratic relation.

STEP 8

First, we expand the quadratic relation to identify the coefficient of $x^2$ .
$y = -3(x-5)(x+7)$

STEP 9

Apply the distributive property (FOIL method) to expand the quadratic relation.
$y = -3(x^2 + 7x - 5x - 35)$

STEP 10

Combine like terms in the expansion.
$y = -3(x^2 + 2x - 35)$

STEP 11

Distribute the -3 across the terms inside the parentheses.
$y = -3x^2 - 6x + 105$

STEP 12

The coefficient of the $x^2$ term is -3.

STEP 13

Since the coefficient of the $x^2$ term is negative, the parabola opens downward.
The zeros of the relation are $x = 5$ and $x = -7$ , and the parabola opens downward.

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Find the zeros of the quadratic relation $y=-3(x-5)(x+7)$ and determine if the parabola opens upward or downward.

STEP 1

Assumptions
1. The quadratic relation is given by $y=-3(x-5)(x+7)$ .
2. Zeros of the relation are the values of $x$ for which $y=0$ .
3. The direction in which the parabola opens is determined by the sign of the coefficient of the $x^2$ term.

STEP 2

To find the zeros of the relation, we set $y$ to zero and solve for $x$ .
$0 = -3(x-5)(x+7)$

STEP 3

Since the product of the two factors is zero, we can set each factor equal to zero and solve for $x$ .
$(x-5) = 0 \quad \text{or} \quad (x+7) = 0$

STEP 4

Solve the first equation for $x$ .
$x - 5 = 0 \Rightarrow x = 5$

STEP 5

Solve the second equation for $x$ .
$x + 7 = 0 \Rightarrow x = -7$

STEP 6

The zeros of the relation are $x = 5$ and $x = -7$ .

STEP 7

To determine the direction in which the parabola opens, we look at the coefficient of the $x^2$ term in the quadratic relation.

STEP 8

First, we expand the quadratic relation to identify the coefficient of $x^2$ .
$y = -3(x-5)(x+7)$

STEP 9

Apply the distributive property (FOIL method) to expand the quadratic relation.
$y = -3(x^2 + 7x - 5x - 35)$

STEP 10

Combine like terms in the expansion.
$y = -3(x^2 + 2x - 35)$

STEP 11

Distribute the -3 across the terms inside the parentheses.
$y = -3x^2 - 6x + 105$

STEP 12

The coefficient of the $x^2$ term is -3.

STEP 13

Since the coefficient of the $x^2$ term is negative, the parabola opens downward.
The zeros of the relation are $x = 5$ and $x = -7$ , and the parabola opens downward.

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Find the zeros of the quadratic relation y=−3(x−5)(x+7)y=-3(x-5)(x+7)y=−3(x−5)(x+7) and determine if the parabola opens upward or downward.

STEP 1

Assumptions1. The quadratic relation is given by y=−3(x−5)(x+7)y=-3(x-5)(x+7)y=−3(x−5)(x+7).2. Zeros of the relation are the values of xxx for which y=0y=0y=0.3. The direction in which the parabola opens is determined by the sign of the coefficient of the x2x^2x2 term.

STEP 2

To find the zeros of the relation, we set yyy to zero and solve for xxx.0=−3(x−5)(x+7)0 = -3(x-5)(x+7)0=−3(x−5)(x+7)

STEP 3

Since the product of the two factors is zero, we can set each factor equal to zero and solve for xxx.(x−5)=0or(x+7)=0(x-5) = 0 \quad \text{or} \quad (x+7) = 0(x−5)=0or(x+7)=0

STEP 4

Solve the first equation for xxx.x−5=0⇒x=5x - 5 = 0 \Rightarrow x = 5x−5=0⇒x=5

STEP 5

Solve the second equation for xxx.x+7=0⇒x=−7x + 7 = 0 \Rightarrow x = -7x+7=0⇒x=−7

STEP 6

The zeros of the relation are x=5x = 5x=5 and x=−7x = -7x=−7.

STEP 7

To determine the direction in which the parabola opens, we look at the coefficient of the x2x^2x2 term in the quadratic relation.

STEP 8

First, we expand the quadratic relation to identify the coefficient of x2x^2x2.y=−3(x−5)(x+7)y = -3(x-5)(x+7)y=−3(x−5)(x+7)

STEP 9

Apply the distributive property (FOIL method) to expand the quadratic relation.y=−3(x2+7x−5x−35)y = -3(x^2 + 7x - 5x - 35)y=−3(x2+7x−5x−35)

STEP 10

Combine like terms in the expansion.y=−3(x2+2x−35)y = -3(x^2 + 2x - 35)y=−3(x2+2x−35)

STEP 11

Distribute the -3 across the terms inside the parentheses.y=−3x2−6x+105y = -3x^2 - 6x + 105y=−3x2−6x+105

STEP 12

The coefficient of the x2x^2x2 term is -3.

STEP 13

Since the coefficient of the x2x^2x2 term is negative, the parabola opens downward.The zeros of the relation are x=5x = 5x=5 and x=−7x = -7x=−7, and the parabola opens downward.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Find the zeros of the quadratic relation $y=-3(x-5)(x+7)$ and determine if the parabola opens upward or downward.

Assumptions
1. The quadratic relation is given by $y=-3(x-5)(x+7)$ .
2. Zeros of the relation are the values of $x$ for which $y=0$ .
3. The direction in which the parabola opens is determined by the sign of the coefficient of the $x^2$ term.

To find the zeros of the relation, we set $y$ to zero and solve for $x$ .
$0 = -3(x-5)(x+7)$

Since the product of the two factors is zero, we can set each factor equal to zero and solve for $x$ .
$(x-5) = 0 \quad \text{or} \quad (x+7) = 0$

Solve the first equation for $x$ .
$x - 5 = 0 \Rightarrow x = 5$

Solve the second equation for $x$ .
$x + 7 = 0 \Rightarrow x = -7$

The zeros of the relation are $x = 5$ and $x = -7$ .

To determine the direction in which the parabola opens, we look at the coefficient of the $x^2$ term in the quadratic relation.

First, we expand the quadratic relation to identify the coefficient of $x^2$ .
$y = -3(x-5)(x+7)$

Apply the distributive property (FOIL method) to expand the quadratic relation.
$y = -3(x^2 + 7x - 5x - 35)$

Combine like terms in the expansion.
$y = -3(x^2 + 2x - 35)$

Distribute the -3 across the terms inside the parentheses.
$y = -3x^2 - 6x + 105$

The coefficient of the $x^2$ term is -3.

Since the coefficient of the $x^2$ term is negative, the parabola opens downward.
The zeros of the relation are $x = 5$ and $x = -7$ , and the parabola opens downward.